OpenJudge/Poj 1915 Knight Moves

1.链接地址：

http://bailian.openjudge.cn/practice/1915

http://poj.org/problem?id=1915

2.题目：

总Time Limit:

1000ms

Memory Limit:

65536kB

Description

Background
Mr Somurolov, fabulous chess-gamer indeed,
asserts that no one else but him can move knights from one position to
another so fast. Can you beat him?
The Problem
Your task is
to write a program to calculate the minimum number of moves needed for a
knight to reach one point from another, so that you have the chance to
be faster than Somurolov.
For people not familiar with chess, the possible knight moves are shown in Figure 1.

Input

The input begins with the number n of scenarios on a single line by itself.
Next
follow n scenarios. Each scenario consists of three lines containing
integer numbers. The first line specifies the length l of a side of the
chess board (4 <= l <= 300). The entire board has size l * l. The
second and third line contain pair of integers {0, …, l-1}*{0, …,
l-1} specifying the starting and ending position of the knight on the
board. The integers are separated by a single blank. You can assume that
the positions are valid positions on the chess board of that scenario.

Output

For each scenario of the input you have to calculate the minimal
amount of knight moves which are necessary to move from the starting
point to the ending point. If starting point and ending point are
equal,distance is zero. The distance must be written on a single line.

Sample Input

3
8
0 0
7 0
100
0 0
30 50
10
1 1
1 1

Sample Output

5
28
0

Source

TUD Programming Contest 2001, Darmstadt, Germany

3.思路：

4.代码：

 #include<stdio.h>
 #include<iostream>
 #include<queue>
 using namespace std;
 typedef struct
 {
     int row;
     int col;
     //int step;
 }data;
 int a[][];
 int colStep[]={,,-,-,,,-,-};
 int rowStep[]={,,-,-,-,-,,};
 bool in(int row,int col,int size)
 {
     if(row<||row>=size) return false;
     if(col<||col>=size) return false;
     return true;
 }
 void initArray(int size)
 {
     int i,j;
     for(i=;i<size;i++)
     {
         for(j=;j<size;j++)
         {
             a[i][j]=-;
         }
     }
 }
 int f(int i,int j,int m,int n,int size)
 {
     queue<data> q;
     int k;
     int newRow,newCol;
     data start,aData;
     start.row=i;
     start.col=j;
     initArray(size);
     //start.step=1;
     a[i][j]=;
     q.push(start);
     while(!q.empty())
     {
         aData=q.front();
         if(aData.row==m&&aData.col==n)
         {
             return a[aData.row][aData.col];
         }
         else
         {
             for(k=;k<;k++)
             {
                 newRow=aData.row+rowStep[k];
                 newCol=aData.col+colStep[k];
                 if(in(newRow,newCol,size))
                 {
                     if(a[newRow][newCol]==-)
                     {
                         data newData;
                         newData.col=newCol;
                         newData.row=newRow;
                         a[newRow][newCol]=a[aData.row][aData.col]+;
                         q.push(newData);
                     }
                 }
             }
         }
         q.pop();
     }
     return ;
 }
 int main()
 {
     int sum,k;
     int i,j,m,n,size;
     cin>>sum;
     for(k=;k<sum;k++)
     {
         cin>>size>>i>>j>>m>>n;
         cout<<f(i,j,m,n,size)<<endl;
     }
     return ;
 }